2.6_problem_and_solution (详细版)

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problem-solutionPPT课件

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Write a composition , following the problem-solution-evaluation pattern.
Topic:
How to …… (eg. keep healthy)
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Development
a general statement
problem
solution
evaluation
Sample
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Writing—Structured Writing
Sample
a general statement problem
Parents should be the main solution to children’s lack of physical exercise. Clearly children are spending too many hours watching TV. Their lack of exercise could lead to health problems. If parents set a good example and take exercise with their children, it would seem a natural part of life. Parents can motivate their children to be active from an early age.
To be continued
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Writing—Structured Writing
so as to suit the on-going development of the national economy. Second, the government should provide college graduates with more opportunities to develop new skills. The third solution, I think, is to adopt strong measures of birth control because the greater the population, the less the opportunity for everyone to get a job.s Nhomakorabealution

2017 AMC 12B Problem and Solution

2017 AMC 12B2017 AMC 12A problems and solutions. The test was held on February 7, 2017.Problem 1Kymbrea's comic book collection currently has comic books in it, and she is adding to her collection at the rate of comic books per month. LaShawn's collection currently has comic books in it, and he is adding to his collection at the rate of comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?SolutionKymbrea has comic books initially and every month, she adds two. This can be represented as where x is the number of months elapsed. LaShawn's collection, similarly, is . To find when LaShawn will have twice the number of comic books asKymbrea, we solve for x with the equation and get .Problem2Real numbers , , and satisfy the inequalities , , and . Which of the following numbers is necessarily positive?SolutionNotice that must be positive because . Therefore the answer is . The other choices:As grows closer to , decreases and thus becomes less than .can be as small as possible (), so grows close to as approaches .For all , , and thus it is always negative.The same logic as above, but when this time.Problem3Supposed that and are nonzero real numbers such that . What is the value of ?SolutionsSolution 1Rearranging, we find , or . Substituting, we canconvert the second equation into .Solution 2Substituting each and with , we see that the given equation holds true,as . Thus,Problem4Samia set off on her bicycle to visit her friend, traveling at an average speed of kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at kilometers per hour. In all it took her minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?Solution 1Let's call the distance that Samia had to travel in total as , so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they areboth , or .She bikes at a rate of kph, so she travels the distance she bikes in hours. She walks at a rate of kph, so she travels the distance she walks in hours.The total timeis . This is equal to of an hour. Solving for , we have:Since is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about .Problem 5The data set has median , firstquartile , and third quartile . An outlier in a data set is a value that is more than times the interquartile range below the first quartle () or more than times the interquartile range above the third quartile (), where the interquartile range is definedas . How many outliers does this data set have?SolutionThe interquartile range is defined as , which is . times this value is , so all values more than below = is an outlier. The only one that fits this is . All values more than above = are also outliers, of whichthere are none so there is onlyProblem 6The circle having and as the endpoints of a diameter intersects the -axis at a second point. What is the -coordinate of this point?SolutionBecause the two points are on a diameter, the center must be halfway between them at the point (4,3). The distance from (0,0) to (4,3) is 5 so the circle has radius 5. Thus, the equationof the circle is .To find the x-intercept, y must be 0, so ,so , , .Problem 7The functions and are periodic with least period . What is the least period of the function ?The function is not periodic. Solutionhas values at its peaks and x-intercepts. Increase them to . Then we plug theminto . and .So, isSolution by TheUltimate123 (Eric Shen)Solution IIStart by noting that . Then realize that under this function the negative sine values yield the same as their positive value, so take the absolute value of the sine function to get the new period. This has period , so the answer is surprisingly !!!!Problem 8The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?Solution 1: Cross-MultiplicationLet be the short side of the rectangle, and be the long side of the rectangle. The diagonal,therefore, is . We can get the equation . Cross-multiplying, weget . Squaring both sides of the equation, we get , which simplifies to . Solving for a quadratic in , using the quadratic formulawe get which gives us . We know that the square of the ratio must be positive (the square of any real number is positive), so the solutionis .Solution by: vedadehhcSolution 2: SubstitutionSolution by HydroQuantumLet the short side of the rectangle be and let the long side of the rectangle be . Then, the diagonal, according to the Pythagorean Theorem, is . Therefore, we can write the equation:.We are trying to find the square of the ratio of to . Let's let our answer, , be . Then, squaring the above equation,.Thus, .Multiplying each side of the equation by ,.Adding each side by ,.Solving for using the Quadratic Formula,.Since the ratio of lengths and diagonals of a rectangle cannot be negative, and ,the symbol can only take on the . Therefore,.Problem 9A circle has center and has radius . Another circle has center andradius . The line passing through the two points of intersection of the two circles has equation . What is ?Solution 1The equations of the two circlesare and . Rearrange themto and , respectively. Their intersection points are where these two equations gain equality. The two points lie on the line with the equation . We cansimplify this like thefollowing.. Thus, .Solution by TheUltimate123 (Eric Shen)Solution 2: Shortcut with right trianglesNote the specificity of the radii, and , and that specificity is often deliberately added tosimplify the solution to a problem.One may recognize as the hypotenuse of the right triangle and as thehypotenuse of the right triangle with legs and . We can suppose that the legs of these triangles connect the circles' centers to their intersection along the gridlines of the plane.If we suspect that one of the intersections lies units to the right of and units above the center of the first circle, we find the point , which is infact unit to the left of and units below the center of the second circle at .Plugging into gives us .A similar solution uses the other intersection point, .Problem10At Typico High School, of the students like dancing, and the rest dislike it. Of those who like dancing, say that they like it, and the rest say that they dislike it. Of those who dislike dancing, say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?SolutionWLOG, let there be students. of them like dancing, and do not. Of those who like dancing, , or of them say they dislike dancing. Of those who dislike dancing, ,or of them say they dislike it. Thus,Problem11Call a positive integer if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. Forexample, , , and are monotonous, but , , and are not. How many monotonous positive integers are there?Solution 1Case 1: monotonous numbers with digits in ascending orderThere are ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. Also, (the empty set) isn't included because it doesn't generate a number. The sum isequivalent toCase 2: monotonous numbers with digits in descending orderThere are ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros.However, (the empty set) still isn't included because it doesn't generate a number. The sumis equivalent to We discard the number 0 since it is not positive. Thus there are here.Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are monotonous numbers.Solution 2Like Solution 1, divide the problem into an increasing and decreasing case:Case 1: Monotonous numbers with digits in ascending order.Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get monotonous numbers for this case.Case 2: Monotonous numbers with digits in descending order.This time, we arrange all 10 digits in decreasing order and repeat the process tofind ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get monotonous numbers for this case.At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.Thus our final answer is .Problem12What is the sum of the roots of that have a positive real part?Solution 1The root of any polynomial of the form will have all of it roots will havemagnitude and be the vertices of a regular -gon in the complex plane (This concept is known as the Roots of Unity). For the equation , it is easy tosee and as roots. Graphing these in the complex plane, we have four vertices of a regular dodecagon. Since the roots must be equally spaced, besides , there are fourmore roots with positive real parts lying in the first and fourth quadrants. We also know that the angle between these roots is . We only have to find the real parts of the roots lying in the first quadrant, because the imaginary parts would cancel out with those from the fourth quadrant. We have two triangles (the triangles formed by connecting the origin to the roots, and dropping a perpendicular line from each root to the real-axis), both withhypotenuse . This means that one has base and the other has base . Adding these and multiplying by two, we get the sum of the four roots as . However, wehave to add in the original solution of , so the answer is .Solution by vedadehhcSolution 2has a factor of , so we need to remember to multiply our solution below, usingthe Roots of Unity. We notice that the sum of the complex parts of all these roots is , because the points on the complex plane are symmetric. The rootswith are and by the Roots of Unity. Their real partsare and . Their sumis . But, remember to multiply by . Theanswer is .Solution by TheUltimate123 (Eric Shen)Problem 13In the figure below, of the disks are to be painted blue, are to be painted red, and is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?SolutionLooking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green disk is in a corner. This yields possible arrangements for the blue disks and red disks in the remaining available slots. Now, consider the case that the green disk is on an edge. This yields more possiblearrangements for the blue disks and red disks in the remaining available slots. Thus, our answer isProblem14An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches?SolutionSolution 1:The top cone has radius 2 and height 4 so it has volume .The frustum is made up by taking away a small cone of radius 1, height 4 from a large cone of radius 2, height 8, so it has volume .Adding, we get .Solution by: SilverLionSolution 2:Find the area of the cone with the method in Solution 1. The area of the frustrumisAdding, we getProblem 15Let be an equilateral triangle. Extend side beyond to a point sothat . Similarly, extend side beyond to a point so that , and extend side beyond to a point so that . What is the ratio of the area of to the area of ?Solution 1: Law of CosinesSolution by HydroQuantumLet .Recall The Law of Cosines. Letting ,Since both and are both equilateral triangles, they must be similar dueto similarity. This means that .Therefore, our answer is .Solution 2: InspectionNote that the height and base of are respectively 4 times and 3 times thatof . Therefore the area of is 12 times that of .By symmetry, . Adding the areas of these three triangles and for the total area of gives a ratio of ,or .Solution 3: CoordinatesFirst we note that due to symmetry. WLOG,let and Therefore, . Using the conditionthat , we get and . It is easy to checkthat . Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio isProblem 16The number has over positive integer divisors. One of them is chosen at random. What is the probability that it is odd?SolutionIf a factor of is odd, that means it contains no factors of . We can find the number of factors of two in by counting the number multiples of , , , and that are less than or equal to .After some quick counting we find that this number is . If the prime factorization of has factors of , there are choices for each divisor for how many factors of should be included (to inclusive). The probability that a randomlychosen factor is odd is the same as if the number of factors of is which is . Solution by: vedadehhcSolution 2We can write as its prime factorization:Each exponent of these prime numbers are one less than the number of factors at play here. This makes sense; is going to have factors: , and the other exponents will behave identically.In otherwords, has factors.We are looking for the probability that a randomly chosen factor of will be odd--numbers that do not contain multiples of as factors.From our earlier observation, the only factors of that are even are ones with at least one multiplier of , so our probability of finding an odd factor becomes the following:Problem 17A coin is biased in such a way that on each toss the probability of heads is and the probability of tails is . The outcomes of the tosses are independent. A player has the choiceof playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?The probability of winning Game A is less than the probability of winning Game B.The probability of winning Game A is less than the probability of winning Game B.The probabilities are the same.The probability of winning Game A is greater than the probability of winning Game B.The probability of winning Game A is greater than the probability of winning Game B. SolutionThe probability of winning Game A is the sum of the probabilities of getting three tails andgetting three heads which is . The probability of winning Game B is the sum of the probabilities of getting two heads and getting two tailssquared. This gives us . The probability of winningGame A is and the probability of winning Game B is , so the answer isProblem18The diameter of a circle of radius is extended to a point outside the circle sothat . Point is chosen so that and line is perpendicular to line . Segment intersects the circle at a point between and . What is the areaof ?Solution 1Let be the center of the circle. Notethat .However, by Power of apoint ,, so .Now. Since .Solution 2: Similar triangles with Pythagoreanis the diameter of the circle, so is a right angle, and therefore by AA similarity, .Because of this, , so . Likewise, , so .Thus the area of .Solution 3: Similar triangles without PythagoreanOr, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:Draw with on . ... ( ratio applied twice).Problem19Let be the -digit number that is formed by writing the integers from to in order, one after the other. What is the remainder when is divided by ?SolutionWe will consider this number and . By looking at the last digit, it is obvious that the number is . To calculate the number , note thatso it is equivalent toLet be the remainder when this number is divided by . We knowthat and , so by the Chinese remainder theorem,since , ,or . So the answer isProblem 20Real numbers and are chosen independently and uniformly at random from theinterval . What is the probability that , where denotes the greatest integer less than or equal to the real number ?SolutionFirst let us take the case that . In this case, both and lie in the interval . The probability of this is . Similarly, in the casethat , and lie in the interval , and the probabilityis . It is easy to see that the probabilitiesfor for are the infinite geometric series that startsat and with common ratio . Using the formula for the sum of an infinite geometric series, we get that the probability is .Problem 21Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?Solution 1Let us simplify the problem. Since all of Isabella's test scores can be expressed as the sumof and an integer between and , we rewrite the problem into receiving scoresbetween and . Later, we can add to her score to obtain the real answer.From this point of view, the problem states that Isabella's score on the seventh test was . We note that Isabella received integer scores out of to . Since is already given as the seventh test score, the possible scores for Isabella on the other six testsare .The average score for the seven tests must be an integer. In other words, six distinct integers must be picked from set above, and their sum with must be a multiple of . The interval containing the possible sums of the six numbers in S arefrom to . We must now find multiples of from the interval to . There are fourpossibilities: , , , . However, we also note that the sum of the six numbers (besides ) must be a multiple of as well. Thus, is the only valid choice.(The six numbers sum to .)Thus the sum of the six numbers equals to . We apply the logic above in a similar way for the sum of the scores from the first test to the fifth test. The sum must be a multiple of . The possible interval is from to . Since the sum of the five scores must be less than , the only possibilities are and . However, we notice that does not work because the seventh score turns out to be from the calculation. Therefore, the sum of Isabella's scores from test to is . Therefore, her score on the sixthtest is . Our final answer is .Solution 2Let be Isabella's average after tests. , so . The only integer between and that satisfies this condition is . Let be Isabella'saverage after tests, and let be her sixth test score. ,so is a multiple of . Since is the only choice that is a multiple of , the answeris .Solution 3Let be the total sum of Isabella's first five test scores, and let be her score on the sixthtest. It follows that , , and , since at each step, her average score was an integer. Using the lastequivalence, , so we have a system of equivalences for . Solving this using the Chinese Remainder Theorem, weget .Now let's put a bound on . Using the given information that each test score was a distinct integer from to inclusive and that the seventh score was 95, weget . Since , we get . Therefore,The last preparation step will involve calculating all the possible test scores . Heretheyare:. This means that . Note that is not in the previous list because it corresponds to a score of , which we cannot have.We must have , and using the possible values we found for and , the only two that sum to are and . This corresponds to an value of , so the answer is .Problem 22Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?SolutionIt amounts to filling in a matrix. Columns are the random draws each round; rows are the coin changes of each player. Also, let be the number of nonzero elements in .WLOG, let . Parity demands that and must equal or .Case 1: and . There are ways to place 's in , so there are ways.Case 2: and . There are ways to place the in , ways to place the remaining in (just don't put it under the on top of it!), and ways for one of the other two players to draw the green ball. (We know it's green because Bernardo drew the red one.) We can just double to cover the case of , for a total of ways.Case 3: . There are three ways to place the in . Now, there are two cases as to what happens next.Sub-case 3.1: The in goes directly under the in . There's obviously way for that to happen. Then, there are ways to permute the two pairs of in and . (Either the comes first in or the comes first in .)Sub-case 3.2: The in doesn't go directly under the in . There are ways to place the , and ways to do the same permutation as in Sub-case 3.1. Hence, thereare ways for this case.There's a grand total of ways for this to happen, along with total cases. The probabilitywe're asking for is thusSolution 2 (Less Casework)We will proceed by taking cases based on how many people are taking part in this "transaction." We can have 2, 3, or 4 people all giving/receiving coins during the 4 turns. Basically, (like the previous solution), we are thinking this as filling out a 4x2 matrix of letters, where a letter on the left column represents this person gave, and a letter on the right column means this person recieved. We need to make sure that for each person that gave a certain amount, they received in total from other people that same amount, or in other words there are an equal number of A's, B's, C's, and D's on both columns of the matrix.Case 1: people. In this case, we can 4C2 ways to choose the two people, and 6 ways to get order them to get a count of 6 * 6 = 36 ways.Case 2: people. In this case, we have 4*(3C2)*4! = 288 ways to order 3 people.Case 3: people. In this case, we have 3*3*4! = 216 ways to order 4 people.So we have a total of 36+288+126=540 ways to order the four pairs of people. Now we divide this by the total number of ways - (4*3)^4 ( 4 times, 4 ways to choose giver, 3 to choose receiver). So the answer is 5/192.~ccx09 (NOTE: Due to the poor quality of this solution, please PM me and I will explain the numbers, I have some diagrams but I can't show it here)Problem 23The graph of , where is a polynomial of degree , containspoints , , and . Lines , , and intersect the graph again at points , , and , respectively, and the sum of the -coordinates of , , and is 24. What is ?SolutionFirst, we can define , which contains points , ,and . Now we find that lines , , and are defined by theequations , , and respectively. Since we want to find the -coordinates of the intersections of these lines and , we set each of them to ,and synthetically divide by the solutions we already know exist (eg. if we were looking atline , we would synthetically divide by the solutions and , because wealready know intersects the graph at and , which have -coordinates of and ). After completing this process on all three lines, we get that the -coordinates of , ,and are , , and respectively. Adding these together, weget which gives us . Substituting this back into the original equation,we get ,andProblem 24Quadrilateral has right angles at and , , and . There is a point in the interior of such that and the areaof is times the area of . What is ?Solution 1Let , , and . Note that . By the PythagoreanTheorem, . Since , the ratios of sidelengths must be equal. Since , and . Let F be a point on such that is an altitude of triangle . Notethat . Therefore, and . Since and form altitudes of triangles and , respectively, the areas of these triangles can be calculated. Additionally, the area of triangle can be calculated, as it is a right triangle. Solving for each of these yields:Therefore, the answerisSolution 2Draw line through , with on and on , . WLOGlet , , . By weighted average .Meanwhile, .. We obtain , namely .The rest is the same as Solution 1.Solution 3Let . Then from the similar triangles condition, we compute and . Hence, the -coordinate of is just . Since lies on the unit circle, we can compute the coordinate as . By Shoelace, we wantFactoring out denominators and expanding by minors, this is equivalent toThis factorsas , so and so the answer is .ProblemA set of people participate in an online video basketball tournament. Each person may be a member of any number of -player teams, but no two teams may have exactly thesame members. The site statistics show a curious fact: The average, over all subsets of size of the set of participants, of the number of complete teams whose members are among those people is equal to the reciprocal of the average, over all subsets of size of the set of participants, of the number of complete teams whose members are among those people. How many values , , can be the number of participants?SolutionSolution by Pieater314159Let there be teams. For each team, there are different subsets of players including that full team, so the total number of team-(group of 9) pairs isThus, the expected value of the number of full teams in a random set of players isSimilarly, the expected value of the number of full teams in a random set of players isThe condition is thus equivalent to the existence of a positive integer such that。

问题解决(problem solving)

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1. 按照决策模型的步骤，应用简易性\有效性矩阵解决案例问题：10分钟
2. 各组汇报最终解决方案（单个或多个）：5分钟 3. 其他组对汇报组现场反馈：5分钟
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课程安排
概述界定问题问题根源分析解决方案制定跟踪、评估问题解决状态
– 工具介绍
– 工具演练
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总结与寄语
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评估问题解决的工具
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工具演练
评估工具
检查表
图示工具
排列图趋势图
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案例分析
第23页
工具演练
继续讨论案例，选择合适的工具评估课程案例中的解决方案
活动要求：15分钟
1、小组讨论，并使用所选工具对问题解决状态进行评估和跟进：10分钟
2、小组派代表分享评估结果：5分钟
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课程安排
概述界定问题问题根源分析解决方案制定跟踪、评估问题解决状态
Process Check to find the defect
Gather the defect and analyze the raw cause
Switchgear FPY
350 300 250 200 150 100 50 5.7% 141 15.0% 80 5.4% 17 112 9.3% 31 6 Jul FPY-Monthly Aug 12 May Total Panel Qty 333 50.0% 45.0% 40.0% 35.0% 30.0% 25.0% 20.0% 15.0%
2+2=?
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需要研究解决的疑难和矛盾
下雨天，没有雨伞怎么回家？
现状与目标的差距

problems and solutions教案

Teaching Plan for Unit 6 Problems and solutionsTeaching objects:Teaching classroom:Teaching tools: PPT, Video, recordingTeaching content: Unit 6 Problems and solutionsTeaching goals:1.get ss to know different industries of business;2.get ss to know typical problem in certain industries;3.get ss to know Just in Time theory.Teaching stress:Just in Time theoryTeaching difficulty:Just in Time theoryTeaching procedures:Step1: ReviewPut yourselves in others' shoesStep2: Lead-in (by Qs with pictures)1.Which industry do you work in?2. Show different business industries3. The work flow of manufacturing cars of Toyota.4. The problem of Toyota.Step3: Reading1. Read this article taken from a business management journal. Underline the words from column A in exercise 4.raw materials loading bay factory floor storage costs waste improvement defects inventory response time stock2. Read the article again and answer the following questions.(1)What was done at Ford Motor company to eliminate storage costs?(2)what name was given to the JIT concept at the Toyota Motor Corporation?(3)What is the main object to JIT?(4)How many sorts of waste have been eliminated in lean manufacturing? What are they?(5)Which sentence in the text summarises the JIT philosophy?3. Language pointsloading bay: the area where good are delivered or taken awayfactory floor: the area where production takes placeadopt: to take and followeliminate: to get rid of; to removecontinuous improvement:持续改进mystery shopper:神秘顾客zero defects:零缺陷minimum：最小值inventory：a detailed lists of of all goods and materials in stockdefects: imperfections or faultsresponse time: the time from getting the order to producing the products4. Ask student to retell the story of Just in Time.5.Work in groups of four. Decide how the JIT concept could be applied in your company. Then form new groups and compare your ideas.。

proble and solution

• One of the main causes of the problem is that teachers’ salaries are lower than many other jobs. Teachers’ pay has not kept pace with that of other professions, such as law or medicine. In the UK, for example, a doctor with five years’ experience will earn far more than a teacher with the same experience. The solution is for the government to raise teachers’ pay significantly, which would attract more people into the profession.
• 0906: Recent crime figures indicate that juvenile delinquency (youth crime) is on the rise in major cities throughout the world. Discuss what the possible reasons are and how to solve the problem.
Causes
Community expectations expand on this thought process by encouraging criminal behaviour in either subtle or overt ways. Economic, social and even cultural factors can further influence delinquent behaviours. The media also occasionally, if not intentionally, promotes violence, which can impact children's actions as well.

青少版新概念2BUnit19Aproblemwithsquirrels

使用情态动词可以使句子更加委婉、礼貌，也可以强调说话人的观点和态度。
04 练习与活动
词汇练习
总结词
掌握重点词汇
总结词
辨析相似词汇
详细描述
学生需要掌握本单元出现的重点词汇，如 “squirrels”（松鼠）、“problem”（问题）、 “solution”（解决方案）等，通过抄写、造句、翻译等方式加深对这些词汇的理解和记忆。
口语活动
总结词
小组讨论与分享
详细描述
教师可以将学生分成小组，让他们围绕与本单元主题相关的议题进行讨论，如“ 如何平衡人与自然的关系”、“如何解决城市中的动物问题”等。通过小组讨论，培养学生的批判性思维和合作精神，同时促进彼此之间的交流与分享。
05 学习总结
本单元重点回顾
重点语法
学习使用情态动词"can"表示可能性，以及使用"if"引导条件状语从句。
写作练习
根据课文内容，模仿写作关于解决动物问题的短文。
重点词汇
squirrel、problem、 solution、prevent、 solution、trap、harmful、 poisonous、kindness、 repay
课文理解
通过阅读课文，理解关于松鼠问题的解决方案，以及其中涉及的环保和动物保护知识。
过去进行时表示在过去某一时刻正在进行的动作或存在的状态。其基本结构是“was/were + -ing”。
过去进行时可以与when、while等连词连用，表示两个动作同时发生。
被动语态
被动语态表示动作的承受者是主语，而不是动作的执行者。其基本结构是“be + 过去分词”。
在本单元中，被动语态的句子有：“The fire was put out by the firemen.”（火被消防员扑灭了。）和“The book was written by a famous author.”（这本书是一位著名作家写的。）

高考英语常考话题：专题03 阅读理解BC+七选五+语法填空考前必练(1)(解析版)(全国通用)

专题03 阅读理解BC+七选五+短文改错必练（1）距离高考还有一段时间，不少有经验的老师都会提醒考生，愈是临近高考，能否咬紧牙关、学会自我调节，态度是否主动积极，安排是否科学合理，能不能保持良好的心态、以饱满的情绪迎接挑战，其效果往往大不一样。

以下是本人从事10多年教学经验总结出的以下学习资料，希望可以帮助大家提高答题的正确率，希望对你有所帮助，有志者事竟成！养成良好的答题习惯，是决定高考英语成败的决定性因素之一。

做题前，要认真阅读题目要求、题干和选项，并对答案内容作出合理预测;答题时，切忌跟着感觉走，最好按照题目序号来做，不会的或存在疑问的，要做好标记，要善于发现，找到题目的题眼所在，规范答题，书写工整;答题完毕时，要认真检查，查漏补缺，纠正错误。

总之，在最后的复习阶段，学生们不要加大练习量。

在这个时候，学生要尽快找到适合自己的答题方式，最重要的是以平常心去面对考试。

英语最后的复习要树立信心，考试的时候遇到难题要想“别人也难”，遇到容易的则要想“细心审题”。

越到最后，考生越要回归基础，单词最好再梳理一遍，这样有利于提高阅读理解的效率。

另附高考复习方法和考前30天冲刺复习方法。

B（2023·河南·校联考一模）As I stood in line waiting to cash out at my favorite retail store, I studied the customer in front of me. She was young, maybe early twenties. But she already had a tired look to her. Her face looked drawn from exhaustion. Her thin, unbuttoned winter coat had seen those bad days. The items in her cart included the cheapest cuts of meat. Powdered milk, day-old bread, bargain soap, and inexpensive shampoo completed her purchases — well, almost.“Next!” When the cashier finally told her the cost, the woman’s face paled. She opened a purse and began counting small bills and change. It was obvious she didn’t have enough, and she scanned her groceries to see what she could do without. One by one, she removed things, but she still came up short. She continued to discard much-needed goods while the cashier set them aside patiently. The child’s items remained in the cart, however.The woman finally paid, and moved down to bag them. When my turn came, I moved forward and placed several items the woman had taken out, whispering “Separate bags, please” to the cashier. I wondered how I could sneak the extra things into the old cart she’d brought without being caught. My clerk solved the problem by “accidentally” rolling several oranges hard enough on the conveyer belt so that they flew past the lady and onto the floor.“I’ll get those,” the woman offered kindly and ran to pick up the runaway fruit. I quickly put the two extra bags into her cart. Just as I put the second one in, I noticed a twenty-dollar bill peeking out that I had not placed in there.“You put that money in there, didn’t you?” I accused the clerk with a laugh when I went to pay my bill. I was surprised when she shook her head. “No, it was her,” she replied, pointing to another woman.My small, insignificant gesture had rippled to become a larger kindness than I ever could have imagined. 1．What can we infer about the customer in front of “me” in Paragraph 1?A．She is a woman to think of others before herself.B．She pretended to be poor to gain other’s sympathy.C．Everything about her screamed hardship and need.D．Everything she chose wasn’t what she truly needed.2．What does the underlined word “discard” probably mean in Paragraph 2?A．Handle. B．Separate. C．Test out. D．Take out.3．Why did the clerk roll oranges hard enough on the conveyer belt?A．To do “me” a favor.B．To train her skill.C．To show her acting experience. D．To help increase grocery sales.4．What can be the best title for the text?A．Kindness in Giving Creates Love B．What Goes around Comes aroundC．Being Happy Is Enough D．More than a Coincidence【答案】1．C 2．D 3．A 4．A【导语】本文是一篇记叙文。

problem,solution模板

竭诚为您提供优质文档/双击可除problem,solution模板篇一：problemandsolutionproblemandsolutiondoyouknowwhymoreandmoreprogramsofblinddatearedesign edespeciallyforwomeninmoderntimestheyarenotonlyfort hepurposeofimprovingtheratingsofthetelevisionstatio ns.yes,theyaredesignedtopandertothepublic`srequirem ents,especiallythewomen`s.becausetodayoldandunmarri edwomenbecomemoreandmore.inaddition,theyareallricha ndhavetheirownstablecareers.theproblemthathasappear edlonglongagoistheopportunityofjobisnotequaltowomen ,especiallytheyoungandunmarriedwomen.asfarasicansee ,apowerfulpolicywhichcanensureswomen`srightsisthemo steffectivewaytosolvetheproblem.nowascompetitionbecomesmoreandmorefierce,thepressur eofobtainingemploymentismoreandmoreserious.theprobl emarosewhenmanymanagersofcompaniesdecidedtohireless unmarriedwomenforthesakeoftheirownbenefitsandavoidi ngtrouble.intheiropinion,onedaywomenwillbebacktothe irfamilies,getmarriedandalsogivebirthtobabies.attha ttime,theywillstillpaythemsalaryandevenbothertofind anotheronetoreplacetheirplaces,whichareagainstbusin essmen`sthoughts.becauseithinkallbusinessmenputthei rbenefitsfirst.soitresultstoproducetheproblem.however,ifthepowerfulpolicyisputintoeffect,thesitua tionwillbedifferent.womenwillhavetheequalrightstofi ndaproperjob.becauseunfairnessisnotaccepted.also,lessoldandunmar riedwomenwillkeepinthesociety.inordertoimplementthepolicy,theattitudeofgovernment officials,managersandwomenshouldchange.First,manage rsshouldrealizethegoodaspectsofhiringunmarriedwomen .justbecausetheyhavenohusbandsandnochildren,theycanhaveenthusiasmforworkingandworkashardaspossible.the n,womenshouldbebraveenoughtostanduptofightfortheirr ights.onlywhenwomenthemselveswakeupcantheygetequali ty.Finally,thegovernmentmustbeawareoftheimportanceo ffairnessandputanemphasisontheunfairphenomenatoassu rewomen`srights.篇二：雅思考官范文中关于solution,measures的秘密-谢爽新东方在线致赢雅思雅思考官范文中关于solution，measures的秘密新东方在线致赢雅思谢爽雅思考试task2中包括四类大作文，分别是opinion，discussion，problemandsolution,two-partquestion，这其中有一类作文仿佛给了我们充分的自由去阐述自己的观点，这就是其中的problemandsolution的题型。

Unit1FromProblemstoSolutions(1)学案含答案高中英语必修第二册

必修2 Unit 1 Cultural HeritageFrom Problems to Solutions1. 学习并理解定语从句的含义、类别和用法；2. 能够熟练运用限制性定语从句表达较为复杂的概念；3. 能够理解并熟练运用本讲核心词汇，描述自己或他人对非物质文化遗产的看法；Reading and thinkingFrom Problems to SolutionsQ1: What are the “problems” and “solutions” in the passage about?Referent answer:Problems: In the 1950s, the Egyptian government wanted to build a new dam across the Nile which would likely damage a number of temples and destroy cultural relics that were an important part of Egypt’s cultural heritage.Solutions: A committee was established and a proposal was made to save the cultural relics which were taken down piece by piece, and then moved and put back together again in a place where they were safe from the water.Q2: The project mentioned in the passage not only took a long time but also cost much money. Do you think it’s necessary? Why?A: I think it is necessary. Because cultural relics are legacies from our ancestors which show us their wisdom and creativity.B: It is quite necessary. Cultural relics provide scientific basis for archaeology which is the study of the societies and peoples of the past.C: It is necessary. Cultural relics could help us to strengthen our national sense of pride and confidence.Water from the dam would likely damage a number of temples and destroy cultural relics that were an important part of Egypt’s cultural heritage.After listening to the scientists who had studied the problem, and citizens who lived near the dam, the government turned to the United Nations for help in 1959.Temples and other cultural sites were taken down piece by piece, and then moved and put back together again in a place where they were safe from the water.Not only had the countries found a path to the future that did not run over the relics of the past, but they had also learnt that it was possible for countries to work together to build a better tomorrow.1. 准确识别定语从句错误：It is known to everybody, the moon travels around the earth once every month.解析：逗号前后为两个简单句，但是并无连词连接，故错误。

Solution_中文软件操作说明书

阿什泰克GPS单双频接收机后处理软件Ashtech Solutions用户操作指南天测企业集团客户服务中心2001年12月北京Solution软件操作说明Solutions软件适用于单频、双频GPS数据基线解算，平差处理。

一、软件安装1、Solutions软件的安装1. 启动Windows，如果Windows已在运行中，应关闭其他应用运行项目。

2. 将光盘插入光驱。

3. 在我的电脑中，打开光驱。

4. 选择Ashtech Solutions文件夹,先安装英文软件。

5. 双击SETUP文件，按照提示安装。

6. 完成安装后，在程序管理器中双击Solution图标，即可启动Solution。

7. 汉化软件安装，在光盘中选择Solutions中文版文件夹，插入启动软盘，双击SETUP文件，按照提示安装。

2、Rinex安装1. 将光盘放入光驱。

2. 在我的电脑中，打开光驱。

3. 选择Rinex文件夹。

4. 双击SETUP按钮，按照提示安装。

5. 完成安装后，在程序管理器中双击Rinex图标，即可启动Rinex。

二、数据解算方法1、数据下载及转换用Mstar软件中Mcomm下载的数据数据后，利用Rinex Converter转换为Rinex 数据格式。

1.1启动Rinex软件在程序管理器中双击Rinex图标，启动Rinex格式转换软件。

如下图：1.2 Rinex数据转换1.2.1选择Promark to Rinex模块，点击Input dir设置要转换的文件所在路径，再点击Output dir设置转换后文件的存储路径。

1.2.2从Promark file中选择需要转换的CAR文件，用鼠标点击窗口正中的三角形BEGIN按钮，即可转换。

2、数据处理用鼠标双击Solution图标，启动Solutions软件，如下图：2.1 建立新项目2.1.1建立新项目的方式有：●方式之一：同时在键盘上按[Ctrl]+[N]键。

●方式之二：单击工具条的[NEW]钮。

Writing-- problems and solutions(共21张PPT)

9. You could ask the teacher for help. 10.You can read faster by reading word groups. 11.You can learn grammar by taking notes ,doing exercises and reading a lot. Solutions 12.How about keeping a diary in English? 13.What about using an English dictionary? 14 Why don’t you listen to English songs ? 15.Why not talk about it with your parents? 16.Maybe you could write a letter to him. 17.Why not sit down and communicate with your brother?
1You should take your temperature. 2.You should lie down and rest. 3.You should drink some hot tea with honey. 4.You should see a dentist and get an XSolutions ray. 5.You should put some medicine on it. 6.You shouldn’t eat so much next time. 7.You shouldn’t play computer games for too long. 8.You shouldn’t stay up late.
My middle school life is coming to an end .During the three years ,not only did I have happy time, but I had kinds of problems. My biggest problem was I had too mu_ch__homework Homework seemed endless for me .It made me stay up late and feel very tired every day ,but I knew the teachers were expecting us to do better,even the best in my lessons.

《问题分析与解决》学习课件

“快速”制定决策
目的：
我需要作什么样的决定？做这样的决定要达到什么目的？
“快速”制定决策
标准：
要是决策成功，哪些要素或特征是必要的？
“快速”制定决策
备选方案：
我有哪些可选择的方案？哪个备选方案更符合标准？
找出选择
备选方案 # 1 决策标准备选方案 # 2
“快速”制定决策
风险：
有哪些潜在的问题可能会导致被选方案失败？
• 刘百功老师的训练特色：尊重平等、互动高效、精辟点评！ • 刘百功老师的训练效果：启迪感悟、擅修行为、力求结果！
思维
• 提高你的思维质量 • 使你的分析更加准确透明 • 改善沟通和团队合作 • 更主动——对事情更加具有驾驭力 • 有效地使用客观信息、事实和数据 • 提高持续改进工作的速度和有效性
情景分析流程
笼统的问题细化并明确问题进行排序决定分析起点
行动方案
•
•
（谁有问题？/流程专家）
•
•
•
•
总结
●明确四种偏差的类型 ●区分解决问题的不同方法 ●理解比较分析法的基本逻辑 ●运用 “快速” 比较分析技巧，
• 明确问题和机会并对其排序，从而确定处理问题的最佳方法
• 将比较分析法广泛地运用于工作中各种问题的分析和改进
• 通过评估关键性的目标和风险来提高决策和方案的效果
• 通过对改进方案、计划和项目积极主动地分析，预防问题的发生，增加投资的机会
●明确大的、全局性的问题 ●明确问题的种类，并通过提问找到真正的问题 ●尽量排除不清晰之处，将精力集中在重要的事项上
领导者之剑

问题与问题解决过程PPT课件

第27页/共69页
步骤一二
三四
任务
具体工作
备注说明
“为什么”
具体化精确化
找到关键
重新表述
“为什么这个问题该我/我们来解决？”
问题所有权
“为什么这个问题显得很重要？”
重要性
“为什么要在这个时候解决这个问题？”
紧迫性
“为什么要提出这个问题？”
原因与背景
“用什么方式解决问题？”
手段的规定
“什么时间与期限完成？”
时间与期限
“什么地点？场合？范围？”
解决问题的环境
“什么程度？幅度？高度？比例？”
解决问题的目标
“达到什么效果？”
解决问题的目的
“问题的陈述中，哪个/几个词是关键词？”
关键点
“这个/几个关键词可以作怎样的表述？” 重新定义关键词
“解决这个问题的主要障碍是什么？”
突破的目标
“解决这个问题所需要的资源和条件是什么？资源与手段
求突破； • 郭泰先生的问题定义术，因为是关于如何进行企划的，离我们最近，但将
“为什么”的探究放到了最后，有本末倒置的嫌疑。
第21页/共69页
C、杨氏界定问题技术
是什么使得思考成为无效思考？恩？
问题陈述与表述方式；问题的所有权与范围；问题的提出者和倡议者的权威；思考者的动机和能力；环境、过程、方法；前景干扰与背景干扰；解决问题的目标不清楚；评价标准模糊。
B、策略规划解决问题过程
• 1）1963年，头脑风暴法的创始人A. F. Osborn作为BBDO广告公司副总裁，针对广告策划工作的特点，提出解决问题的三个步骤：
寻找事实
寻找构想
第7页/共69页
寻求解答

大学英语精读第三版（董亚芬主编）第一册_Unit1~Unit10_翻译答案

Unit1 翻译1) 史密斯太太对我抱怨说，她经常发现与自己十六岁的女儿简直无法沟通。

Mrs. Smith complained to me that she often found it simply impossible to communicate with her 16-year-old daughter.2) 我坚信，阅读简写的(simplified) 英文小说是扩大我们词汇量的一种轻松愉快的方法。

I firmly believe that reading simplified English novels is an easy and enjoyable way of enlarging our vocabulary.3) 我认为我们在保护环境不受污染(pollution) 方面还做得不够。

I don’t think we’re doing enough to protect our environment from pollution.4) 除了每周写作文外，我们的英语老师还给我们布置了八本书在暑假里阅读。

In addition to/Apart from writing compositions on a weekly basis, our English teacher assigned us eight books to read during the summer vacation.5) 我们从可靠的消息来源获悉下学期一位以英语为母语的人将要教我们英语口语。

We’ve learned from reliable sources that a native English speaker is going to teach us spoken English next term/semester.6) 经常看英语电影不仅会提高你的听力，而且还会帮助你培养说的技能。

Seeing English movies on a regular basis will not only improve your ear, but will also help you build your speaking skills.7) 如果你们对这些学习策略有什么问题，请随便问我。

Problems_and_solutions

Problems with solutions :1. A 1-m 3 tank is filled with a gas at room temperature 20°C and pressure 100 Kpa. How much mass is there if the gas isa) Airb) Neon, orc) Propane ?Given: T=273K; P=100KPa; M air =29; M neon =20; M propane =44;TR M V P m ×××= Kg m air 19.12938314291105=×××= Kg m neon 82.019.12920=×= Kg m propane 806.182.02044=×=2. A cylinder has a thick piston initially held by a pin as shown in fig below. The cylinder contains carbon dioxide at 200 Kpa and ambient temperature of 290 k. the metal piston has a density of 8000 Kg/m 3 and the atmospheric pressure is 101 Kpa. The pin is now removed, allowing the piston to move and after a while the gas returns to ambient temperature. Is the piston against the stops?Schematic:100 mm100 mm 50 mm100 mmPinSolution:Given: P=200kpa;332107858.01.01.04m V gas −×=××=π: T=290 k: V piston =0.785×10-3: m piston = 0.785×10-3×8000=6.28 kgPressure exerted by piston =kpa 78481.048.928.62=××πWhen the metal pin is removed and gas T=290 k33221018.115.01.04m v −×=××=π3310785.01m v −×=kpa p 13318.1785.02002=×=Total pressure due to piston +weight of piston =101+7.848kpa=108.848 paConclusion: Pressure is grater than this value. Therefore the piston is resting against the stops.3. A cylindrical gas tank 1 m long, inside diameter of 20cm, is evacuated and then filledwith carbon dioxide gas at 250c.To what pressure should it be charged if there should be1.2 kg of carbon dioxide?Solution: T= 298 k: m=1.2kg:Mpa p15.212.042984483142.12=××××=π4. A 1-m 3 rigid tank with air 1 Mpa, 400 K is connected to an air line as shown in fig: the valve is opened and air flows into the tank until the pressure reaches 5 Mpa, at which point the valve is closed and the temperature is inside is 450 K.a. What is the mass of air in the tank before and after the process?b. The tank is eventually cools to room temperature, 300 K. what is the pressure inside the tank then?Solution:P=106 Pa: P 2=5×106 Pa: T 1=400K: T 2=450 kKg m 72.840083142911061=×××=Kg m 8.3845083142910562=×××=Mpa P 34.313002983148.38=××=5. A hollow metal sphere of 150-mm inside diameter is weighed on a precision beam balance when evacuated and again after being filled to 875 Kpa with an unknown gas. The difference in mass is 0.0025 Kg, and the temperature is 250c. What is the gas, assuming it is a pure substance?Solution:m=0.0025Kg: P=875×103 Kpa: T= 298 K415.06108752980025.0831433=×××××=πMThe gas will be helium.6.Two tanks are connected as shown in fig, both containing water. Tank A is at 200Kpa,ν=1m3 and tank B contains 3.5 Kg at 0.5 Mp, 4000C. The valve is now opened and the two come to a uniform state. Find the specific volume.Schematic:Known:ν=0.61728m/Kg3X=0.61728*3.5= 2.16 KgFinal volume=2.16+1 =3.16 m3Final volume=2+3.5= 5.5 KgFinal specific volume= 3.16/5.5=0.5745 m3/KgTherefore it is a mixture of steamand water.V=1m3M=2 Kgνf =0.001061m3/Kgνg =0.88573 m3/KgT=4000Cm=3.5 Kgkg74.15745.01minA==16 .27.. The valve is now opened and saturated vapor flows from A to B until the pressure in B Consider two tanks, A and B, connected by a valve as shown in fig. Each has a volume of 200 L and tank A has R-12 at 25°C, 10 % liquid and 90% vapor by volume, while tank B is evacuated has reached that in A, at which point the valve is closed. This process occurs slowly such that all temperatures stay at 25 °C throughout the process. How much has the quality changed in tank A during the process?Solution: Given R-12P= 651.6 KPaνg = 0.02685 m3/Kgνf = 0.763*10-3 m 3/Kg310*763.002.002685.018.0m −+== 6.704 + 26.212= 32.9162037.0916.32704.6x 1==Amount of vapor needed to fill tank B =Kg 448.702685.02.0=Reduction in mass liquid in tank A =increase in mass of vapor in Bm f =26.212 –7.448 =18.76 KgThis reduction of mass makes liquid to occupy = 0.763×10-3 ×18.76 m 3 =0.0143 m 3Volume of vapor =0.2 – 0.0143 =0.1857 LMg =Kg 916.602685.01857.0=2694.076.18916.6916.6x 2=+=Δx. =6.6 %8. A linear spring, F =K s (x-x 0), with spring constant K s = 500 N/m, is stretched until it is 100 mm long. Find the required force and work input.Solution:F=K s (x-x o) x- x 0= 0.1 mK s =500 N/mF= 50 N21W =FS =21×50×0.1 =2.539. A piston / cylinder arrangement shown in fig. Initially contains air at 150 kpa, 400°C. The setup is allowed to cool at ambient temperature of 20°C.a. Is the piston resting on the stops in the final state? What is the final pressure inthe cylinder?W b. That is the specific work done by the air during the process?Schematic:Solution:p 1= 150×103 PaT 1=673 KT 2=293 K221111T V P T V P ×=× 1. If it is a constant pressure process, m A V T T V 87.026732931122=××=×=Since it is less than weight of the stops, the piston rests on stops.2211T V T V = T2 =112T V V × =K 5.33626731=× 2233T p T p = KPa T T P P 6.1305.3362931015032323=××=×=Therefore W =Kg KJ A A /5.9629210150831411015033−=××××××××−10. A cylinder, A cyl = 7.012cm2 has two pistons mounted, the upper one, m p1=100kg, initially resting on the stops. The lower piston, m p2=0kg, has 2 kg water below it, with a spring in vacuum connecting he two pistons. The spring force fore is zero when the lower piston stands at the bottom, and when the lower piston hits the stops the volume is 0.3 m3. The water, initially at 50 kPa, V=0.00206 m3, is then heated to saturated vapor.a.Find the initial temperature and the pressure that will lift the upper piston.b.Find the final T, P, v and work done by the water.Schematic:There are the following stages:(1) Initially water pressure 50 kPa results in some compression of springs.Force = 50×103×7.012×10-4 = 35.06 NSpecific volume of water = 0.00206/2 = 0.00103 m 3/kgHeight of water surface = 410012.700206.0−×= 2.94 mSpring stiffness = m N /925.1194.206.35=(2) As heat is supplied, pressure of water increases and is balanced by spring reaction dueto due to K8. This will occur till the spring reaction= Force due to piston + atm pressure=981+105 × 7.012×10-4 =1051 NThis will result when S =m 134.80925.111051=At this average V= 7.012× 10-4 × 88.134 =0.0618m 3P=410012.71051−×=1.5 Mpa(3) From then on it will be a constant pressure process till the lower piston hits the stopper. Process 2-3At this stage V= 0.3 m 3Specific volume = 0.15 m 3/kgBut saturated vapor specific volume at 1.5 Mpa = 0.13177 m 3/ kgV=0.26354 m 3(4) Therefore the steam gets superheated 3-4Work done = p 2(v 4 –v 2)+ 21(p 2 +p 1) (v 2-v 1) =1.5×106(0.15-0.0618) + 21(1.5×106 +50×103)(0.0618 –0.00103) = 178598.5 J= 179 KJ11. Two kilograms of water at 500 kPa, 20°C are heated in a constant pressure process (SSSF) to 1700°C. Find the best estimate for the heat transfer.Solution:-h1)]Q = m [(h=2[(6456-85)]=12743 KJChart data does not cover the range. Approximately h2= 6456KJ/kg; h1=85 KJ;500 kPa 130°C h=5408.57700°C h=3925.97Δh = 1482.6 kJ/kg262 kJ/kg /100°C12. Nitrogen gas flows into a convergent nozzle at 200 kPa, 400 K and very low velocity. It flows out of the nozzle at 100 kPa, 330 K. If the nozzle is insulated, find the exit velocity.Solution:smhhchhcchch/8.381(210004.342100031.4152222122122222211=−=×−×=−=+=+13. An insulated chamber receives 2kg/s R-134a at 1 MPa, 100°c in a line with a low velocity. Another line with R-134a as saturated liquid, 600c flows through a valve to the mixing chamber at 1 Mpa after the valve. The exit flow is saturated vapor at 1Mpa flowing at 20-m/s. Find the flow rate for the second line.Solution:Q=0; W=0;SFEE = 0=m 3 (h 3)+c 32/2 – (m 1h 1+m 2h 2)m 1=2g/s h 1 (1Mpa, 100°C) = 483.36×103 J/kgm 2=? h 2 (saturated liquid 60°C =287.79×103 J/kg)m 3=? h 3( saturated vapor 1Mpa = 419.54×103 J/kg))287790(4833602240041954023m m +×=⎦⎤⎢⎣⎡+419.74 m 3=966.72+287.79m 21.458m 3 = 3.359+m 2m 3 = 2 +m 20.458m 3 = 1.359m 3= 2.967 kg/s ; m 2 = 0.967 kg/s14. A small, high-speed turbine operating on compressed air produces a power output of 100W. The inlet state is 400 kPa,50°C, and the exit state is 150 kPa-30°C. Assuming the velocities to be low and the process to be adiabatic, find the required mass flow rate of air through the turbine.Solution:W100=SFEE : -100 = [h.m2–h1]h1= 243.Cph2=323.Cp-100 = Cp(243-323).mCp=1.25.m=1.25×10.m-3 kg/s15. The compressor of a large gas turbine receives air from the ambient at 95 kPa,20°C, with a low velocity. At the compressor discharge, air exists at 1.52 MPa, 430°C,with a velocity of 90-m/s. The power input to the compressor is 5000 kW. Determinethe mass flow rate of air through the unit.Solution:Assume that compressor is insulated. Q=0;SFEE: 5000×103 = [1000*430 +.m ]2010002902×−5000= [410 –4.05].m=12.3 kg/s.m16. In a steam power plant 1 MW is added at 700°C in the boiler , 0.58 MW is taken atout at 40°C in the condenser, and the pump work is 0.02 MW. Find the plant thermalefficiency. Assuming the same pump work and heat transfer to the boiler is given, howmuch turbine power could be produced if the plant were running in a Carnot cycle?Solution:0.4 MW694.010233131=−=ηTheoretically 0.694 MW could have been generated. So 0K on Carnot cyclePower= 0.694 W17. A car engine burns 5 kg fuel at 1500 K and rejects energy into the radiator andexhaust at an average temperature of 750 K. If the fuel provides 40000 kJ/kg, what isthe maximum amount of work the engine provide?Solution:W%50121=−=T T T ηW= 20,000*5=105 KJ=100MJ18. At certain locations geothermal energy in underground water is available and usedas the energy source for a power plant. Consider a supply of saturated liquid water at150°C. What is the maximum possible thermal efficiency of a cyclic heat engine usingthe source of energy with the ambient at 20°C? Would it be better to locate a source ofsaturated vapor at 150°C than to use the saturated liquid at 150°C?Solution:%7.30307.04232931maxor =−=η19. An air conditioner provides 1 kg/s of air at 15°C cooled from outside atmospheric air at 35°C. Estimate the amount of power needed to operate the air conditioner. Clearly state all the assumptions made.Solution: assume air to be a perfect gas4.1420288==copW W13904.1420080==20. We propose to heat a house in the winter with a heat pump. The house is to be maintained at 20 0C at all times. When the ambient temperature outside drops at –10 0C that rate at which heat is lost from the house is estimated to be 25 KW. What is the minimum electrical power required to drive the heat pump?Solution:KW WHpcop56.271.92577.930293====21.A house hold freezer operates in room at 20°C. Heat must be transferred from thecold space at rate of 2 kW to maintain its temperature at –30°C. What is thetheoretically smallest (power) motor required to operating this freezer?Solution:kW W cop 41.086.4286.450243====22. Differences in surface water and deep-water temperature can be utilized for powergenetration.It is proposed to construct a cyclic heat engine that will operate nearHawaii, where the ocean temperature is 200C near the surface and 50C at some depth.What is the possible thermal efficiency of such a heat engine?Solution: %529315max ==η]23. We wish to produce refrigeration at –300C. A reservoir, shown in fig is available at200 0C and the ambient temperature is 30 0C. This, work can be done by a cyclic heatengine operating between the 200 0C reservoir and the ambient. This work is used todrive the refrigerator. Determine the ratio of heat transferred from 200 0C reservoir tothe heat transferred from the –300C reservoir, assuming all process are reversible.Solution:Q 1/Q 2=?3594.0=η 05.4=cop3594.0×=Q W05.405.422Q W W Q =×=69.03594.005.4105.63594.02121=×==×Q Q Q Q24. Nitrogen at 600 kPa, 127°C is in a 0.5m3-insulated tank connected to pipe with a valve to a second insulated initially empty tank 0.5 m3. The valve is opened and nitrogen fills both the tanks. Find the final pressure and temperature and the entropy generation this process causes. Why is the process irreversible?Solution:Final pressure = 300 kPaFinal temperature=127 kPa as it will be a throttling process and h is constant. T= constant for ideal gaskgm5.28314287504002883145.0600103=×=×××=Δs for an isothermal process=12lnVVmR=22853145.2m×=514.5 J/k25. A mass of a kg of air contained in a cylinder at 1.5Mpa, 100K, expands in a reversible isothermal process to a volume 10 times larger. Calculate the heat transfer during the process and the change of entropy of the air.Solution:V2= 10V11211ln v v v p W Q == For isothermal process=121ln v v mRT J 66012710ln *1000*298314*1==W=Q for an isothermal process,T Δs=660127;Δs=660J/K26. A rigid tank contains 2 kg of air at 200 kPa and ambient temperature, 20°C. An electric current now passes through a resistor inside the tank. After a total of 100 kJ of electrical work has crossed the boundary, the air temperature inside is 80°C, is this possible?Solution:Q=100*103 JIt is a constant volume process.T mc Q v Λ==2×707×20=83840 JQ given 10,000 Joules only. Therefore not possible because some could have been lost through the wall as they are not insulted.K J T dT mc S v air/93.261293353ln 7032353293=×==Δ∫ K J S sun /3.341293101003−=×−=Δ0<Δ+Δsun systemHence not possible. It should be >=0;27. A cylinder/ piston contain 100 L of air at 110 kPa, 25°C. The air is compressed in reversible polytrophic process to a final state of 800 kPa, 2000C. Assume the heat transfer is with the ambient at 25°C and determine the polytrophic exponent n and the final volume of air. Find the work done by the air, the heat transfer and the total entropy generation for the process.Solution:32233222111022.0473108002981.010110m V V T V p T V p ==××=××==31.1)545.4(273.712212211==⎟⎟⎠⎞⎜⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛×=×γγγγγV V p p V p V pW=J n V p V p 21290131.110022.08001.0101101332211−=−××−××=−−JQ UW Q J U KJ S kg m kgK J T T c V V R S v 5110212901618016180)298473(4.029*******.0/28.13129.02982983141.010110/103298473ln 48.12983141.0022.0ln 298314ln ln 32112−=−=Δ=−=−××=Δ−=Δ=×××=−=×+=+=Δ28. A closed, partly insulated cylinder divided by an insulated piston contains air in one side and water on the other, as shown in fig. There is no insulation on the end containing water. Each volume is initially 100L, with the air at 40°C and the water at 90°C, quality 10 %. Heat is slowly transferred to the water, until a final pressure of 500kPa. Calculate the amount of heat transferred.Solution:State 1:Vair=0.1m3 V water =0.1m 3 Total volume=0.2m 3t air =40°C x=0.1 t water =90°CInitial pressure of air = saturation pressure of water at 90°C = 70.14kPav g /90°C =2.360506m 3/kg v f /90°C =0.0010316m 3/kgV = xv g +(1-x)v f=0.1*2.36056+0.9*0.0010316=0.237m 3/kgV=0.1m 3m water =kg V422.0237.01.0==νState 2:Assume that compression of air is reversible. It is adiabaticγγ2211V p V p =34.11121120246.050014.701.0m p p V V =⎟⎠⎞⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛=γVolume of water chamber =0.2- 0.0246=0.1754m 3 Specific volume =kgm v kg m kPa g /3738.0//416.0422.01754.035003==Therefore steam is in superheated state.。

ProblemSolutions...

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Binary 101110001.0101
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Hexadecimal 171.5 BD.A D6.A F3C7.A
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Σm(3, 5, 6, 7)
Problem Solutions – Chapter 2
b)
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Problem Solutions to Problems Marked With a * in
Logic Computer Design Fundamentals, Ed. 2

problems-and-solutions2

Chapter 66。

1。

Carbon dioxide is diffusing through nitrogen in one direction at atmospheric pressure and 0°C. The mole fraction of CO 2 at point A is 0。

2; at point B , 3 m away ， in the direction of diffusion ， it is 0.02. Diffusivity D is 0.144 cm 2/s 。

The gas phase as a whole is stationary ； that is, nitrogen is diffusing at the same rate as the carbon dioxide ， but in the opposite direction 。

（a ） What is the molal flux of CO 2， in kilogram moles per square meter per hour? （b ） What is the net mass flux, in kilograms per square meter per hour ？ (c ) At what speed, in meters per second ， would an observer have to move from one point to the other so that the net mass flux, relative to him or her ， would be zero ？ (d ） At what speed would the observer have to move so that, relative to him or her, the nitrogen is stationary ？（e ） What would be the molal flux of carbon dioxide relative to the observer under condition (d ）？solution :(a ）from(6.1—8）27308206.01⨯===+RT P c c c M B A from equation(6.1—19)()()h m kmol y y z Dc J N A Ai M A A 244/10388.102.02.027308206.03360010144.0--⨯=-⨯⨯⨯⨯=-== (b ) net mass fluxfor carbon dioxide (molecular weight=44） mass flux of CO 2= 44×1.388×10-4 kg/m 2h for nitrogen (molecular weight=28)mass flux of N 2= 28×1.388×10—4 kg/m 2hso the net mass flux in the direction of CO 2 diffusionm =(44-28) ×1。

Unit6Problemsandsolutions

Banning images of beautiful women in her shops, she promoted health, not beauty.
Expensive.
Unimportant.
●How they By testing animals. By funding research into
Expensive.
Other cosmetic companies
●How they publicize(推广) their products ●What images are used to sell their products ●Packaging
●How they test product safety
T or F? Why?1.If the cosmetic products are harmless, the
company may be charged for millions of dollars.( T )
How?2.MAanlly cosmetic products are tested on animals to
and expensive
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advertising
and later by
campaigns.
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Pictures of youthful girls.
Paபைடு நூலகம்a3
1. When and where was the Body Shop set up? In 1976, in England.